Calculus

Infinite Sequences and Series

Infinite Sequences

A sequence of real numbers \(n\) is a function \(f\left( n \right),\) whose domain is the set of positive integers. The values \({a_n} = f\left( n \right)\) taken by the function are called the terms of the sequence.

The set of values \({a_n} = f\left( n \right)\) is denoted by \(\left\{ {{a_n}} \right\}.\)

A sequence \(\left\{ {{a_n}} \right\}\) has the limit \(L\) if for every \(\varepsilon \gt 0\) there exists an integer \(N \gt 0\) such that if \(n \ge N,\) then \(\left| {{a_n} – L} \right| \le \varepsilon .\) In this case we write:

\[\lim\limits_{n \to \infty } {a_n} = L.\]

The sequence \(\left\{ {{a_n}} \right\}\) has the limit \(\infty\) if for every positive number \(M\) there is an integer \(N \gt 0\) such that if \(n \ge N\) then \({a_n} \gt M.\) In this case we write

\[\lim\limits_{n \to \infty } {a_n} = \infty.\]

If the limit \(\lim\limits_{n \to \infty } {a_n} = L\) exists and \(L\) is finite, we say that the sequence converges. Otherwise the sequence diverges.

The sequence \(\left\{ {{a_n}} \right\}\) is bounded if there is a number \(M \gt 0\) such that \(\left| {{a_n}} \right| \le M\) for every positive \(n.\)

Every convergent sequence is bounded. Every unbounded sequence is divergent.

The sequence \(\left\{ {{a_n}} \right\}\) is monotone increasing if \({a_n} \le {a_{n + 1}}\) for every \(n \ge 1.\) Similarly, the sequence \(\left\{ {{a_n}} \right\}\) is called monotone decreasing if \({a_n} \ge {a_{n + 1}}\) for every \(n \ge 1.\) The sequence \(\left\{ {{a_n}} \right\}\) is called monotonic if it is either monotone increasing or monotone decreasing.

Solved Problems

Click or tap a problem to see the solution.

Example 1

Write a formula for the \(n\)th term of \({a_n}\) of the sequence and determine its limit (if it exists).

Example 2.

We easily can see that \(n\)th term of the sequence is given by the formula \({a_n} = {\large\frac{{{{\left( { – 1} \right)}^{n – 1}}n}}{{{2^{n – 1}}}}\normalsize}.\) Since \( – n \le {\left( { – 1} \right)^{n – 1}}n \le n,\) we can write: